/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A camera lens \((n=1.50)\) is co... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 19

A camera lens \((n=1.50)\) is coated with a thin film of magnesium fluoride \((n=1.38)\) of thickness \(90.0 \mathrm{nm}\) What wavelength in the visible spectrum is most strongly transmitted through the film?

Short Answer

Expert verified
Answer: The most strongly transmitted wavelength through the magnesium fluoride film is approximately 686 nm.

Step by step solution

01

Identify relevant formulas and concepts

In a thin film of material, light waves will interact with each other, causing interference. For constructive interference to occur, the additional path length travelled by light waves through the film must be an integer multiple of the wavelength. The general formula for constructive interference in a thin film is given by: \(2nt=m位'\), where \(n\) is the refractive index of the film material, \(t\) is the thickness of the film, \(m\) is an integer multiple, and \(位'\) is the wavelength of light in the film. The relationship between the wavelength of light in the film \(位'\) and the wavelength of light in air (or vacuum) \(位\) is given by \(位'=\frac{位}{n}\).
02

Calculate the path length

We are given the refractive index of magnesium fluoride (\(n = 1.38\)) and the thickness of the film (\(t = 90.0\,\text{nm}\)). Using these values, we can calculate the path length, \(2nt\), by simply multiplying them together, and then multiplying by 2. Path length, \(2nt = 2 脳 1.38 脳 90.0\,\text{nm} = 248.4\,\text{nm}\).
03

Determine the constructive interference condition

Recall that the constructive interference condition is given by \(2nt=m位'\). We already have the path length \(2nt=248.4\,\text{nm}\), so we will use \(位'=\frac{位}{n}\) to rewrite the condition in terms of the wavelength of light in air. We get \(248.4\,\text{nm} = m\frac{位}{1.38}\).
04

Choose the visible wavelength range

The visible spectrum of light consists of wavelengths from about \(380\,\text{nm}\) to \(750\,\text{nm}\). We need to find the wavelength in this range that is most strongly transmitted through the film.
05

Determine the integer multiple, m

In order to find the most strongly transmitted wavelength, we need to find the integer multiple, \(m\), that satisfies the constructive interference condition (\(248.4\,\text{nm} = m\frac{位}{1.38}\)) for a wavelength within the visible spectrum. We can solve for \(m\) and then for \(位\): \(m=\frac{248.4\,\text{nm}脳1.38}{位}\). In this case, we will use an iterative method, testing values of \(m\) until we find the one that will give us a wavelength in the visible range.
06

Calculate the most strongly transmitted wavelength

By using different integer values for \(m\) and calculating the corresponding wavelengths \(位\), we find that when \(m=2\), the wavelength is within the visible range: \(位 = \frac{248.4\,\text{nm}脳1.38}{2} 鈮� 686\,\text{nm}\). This wavelength is within the visible range (between \(380\,\text{nm}\) and \(750\,\text{nm}\)) and close to the red end of the spectrum. So the most strongly transmitted wavelength through the magnesium fluoride film is approximately \(686\,\text{nm}\).

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Most popular questions from this chapter

In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).
A mica sheet \(1.00 \mu \mathrm{m}\) thick is suspended in air. In reflected light, there are gaps in the visible spectrum at \(450,525,\) and $630 \mathrm{nm} .$ Calculate the index of refraction of the mica sheet.
A thin layer of an oil \((n=1.60)\) floats on top of water \((n=1.33) .\) One portion of this film appears green \((\lambda=510 \mathrm{nm})\) in reflected light. How thick is this portion of the film? Give the three smallest possibilities.
Show that the interference fringes in a double-slit experiment are equally spaced on a distant screen near the center of the interference pattern. [Hint: Use the small angle approximation for \(\boldsymbol{\theta} .]\)
A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).
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